Question

A bar magnet has a magnetic moment equal to $5\times {10}^{-5}\Omega m.$ It is suspended in magnetic field, which has a magnetic induction $B=8\pi \times {10}^{-4}T.$ Then, the magnet vibrates with period of vibration equal to 15 s. Then the moment of inertia of magnetic well be :

• A. $7.16\times {10}^{-7}kg-m^2$
• B. $14.32\times {10}^{7}kg-m^2$
• C. $3.58\times {10}^{-7}kg-m^2$
• D. None of these.

Answer 1

The correct option is A $7.16\times {10}^{-7}kg-m^2$

Magnetic moment of bar magnet

$M=5\times {10}^{-5}\Omega -m$

$B=8\pi \times {10}^{-4}T$

$T=2\pi \sqrt{\frac{l}{MB}}$

$T^2=4{\pi }^2\frac{l}{MB}$

$l=\frac{T^2}{4p{i}^2}MB$

$=\frac{(15{)}^2}{4\times (3.14{)}^2}\times 5\times {10}^{-5}\times 8\pi \times {10}^{-4}$

$=7.16\times {10}^{-7}kg-m^2$